Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

There are n different size pairs of shoes in the box. One [#permalink]

Show Tags

02 May 2008, 10:04

1

This post received KUDOS

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<2n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?

Last edited by lexis on 06 May 2008, 23:20, edited 1 time in total.

There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?

I get nC1 * 2n-2C2k-2 / 2nC2k = (2*k^2-k)/(2*n-1) .

There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?

I get nC1 * 2n-2C2k-2 / 2nC2k = (2*k^2-k)/(2*n-1) .

u mean nC1=C(n,1)?

U should explain how did U get it. As I see, your answer is not correct.

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards
_________________

mates, please visit my profile and leave comments http://gmatclub.com/forum/johnlewis1980-s-profile-feedback-is-more-than-welcome-80538.html

I'm not linked to GMAT questions anymore, so, if you need something, please PM me

I'm already focused on my application package

My experience in my second attempt http://gmatclub.com/forum/p544312#p544312 My experience in my third attempt http://gmatclub.com/forum/630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards

WELL, your answer is not correct. -----------

Let me explain more about this statement: For example, there are 5 pairs of shoes A,B,C,D,E probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?

Mean: Let A1, A2 is correct pair (pretend) C1,D2 is incorrect pair or C2, E2 or... (pretend)

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards

WELL, your answer is not correct. -----------

Let me explain more about this statement: For example, there are 5 pairs of shoes A,B,C,D,E probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?

Mean: Let A1, A2 is correct pair (pretend) C1,D2 is incorrect pair or C2, E2 or... (pretend)

Hope it helps you solve the general puzzle.

@ RyanDe680: Your avatar is so interesting.

I'm afraid so

but why?

Don't we agree in the probability to take one complete pair of shoes? i.e. to take the right shoe once you've already take one out?

For me: 1/(2n-1)

Explanation: you take one shoe out, therefore, just 2n-1 shoes remain in the box. The probability to take the right one off is 1/(2n-1), doesn't it?

I need to improve my statistic skill.

Thanks for the explanation
_________________

mates, please visit my profile and leave comments http://gmatclub.com/forum/johnlewis1980-s-profile-feedback-is-more-than-welcome-80538.html

I'm not linked to GMAT questions anymore, so, if you need something, please PM me

I'm already focused on my application package

My experience in my second attempt http://gmatclub.com/forum/p544312#p544312 My experience in my third attempt http://gmatclub.com/forum/630-q-47-v-28-engineer-non-native-speaker-my-experience-78215.html#p588275

You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.

I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.

You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.

alexsmith wrote:

I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.

alex, please, doesn't think that all here are fools. Be frank and post advertisement in appropriate threads.
_________________

I'm just giving my views, I'm not advertising. Is it so that if someone gives their views about a site which I used for my preparation of GMAT is called advertisement?? I benefited from this site and I want everyone else to also. I'll keep giving my views about the source what I used for my GMAT entrance and will ask all to just try it once. I'm sure you will get the results that you are dreaming for. Thats all I want to say.

This isn't an official question, right? The variables are awkwardly defined.

Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.

Congratulation! You're correct. Your math skill is very good! The tricky is to require ONLY ONE pair of shoes be same size. To solve it, we separate two sequences: 1st: There is NO pair of shoes 2nd: There is ONLY ONE pair of shoes be same size ==> Combine: ONLY ONE pair of shoes be same size + NO pair of shoes in rest shoes (2n-2)